Let $X$ be a random variable following distribution function (i.e., generalized Pareto distribution)
$$ F_{\gamma, \sigma}(x) = 1-\left( 1+\frac{\gamma x}{\sigma} \right)^{-\frac{1}{\gamma}}, $$
where $\sigma>0, \gamma\in\mathbb R$. In addition,$x\in[0, \infty)$ if $\gamma\geq 0$ and $x\in\left[0, -\frac{\sigma}{\gamma}\right]$ if $\gamma<0$.
It is claimed that $U:=\left( 1+\frac{\gamma X}{\sigma} \right)^{-\frac{1}{\gamma}}$ is uniformly distributed on $(0,1)$. To check this claim it is sufficient to show that the distribution function for $U$ is $F_U(u)=u$.
$$ F_U(u) = \mathbb P(U\leq u) = \mathbb P \left[\left( 1+\frac{\gamma X}{\sigma} \right)^{-\frac{1}{\gamma}}\leq u\right] = F_{\gamma, \sigma}\left[ \frac{\sigma}{\gamma}\left( u^{-\gamma}-1 \right) \right] = 1-u. $$
From the above calculation, it seems that the claim does NOT check out. Is there any explanation for the claim, please? Thank you!
The inequalities are not conserved in these manipulations.
If $\gamma>0$, then \begin{align*} F_U(u) = \mathbb P(U\leq u) = \mathbb P \left[\left( 1+\frac{\gamma X}{\sigma} \right)^{-\frac{1}{\gamma}}\leq u\right] &=\mathbb P \left[1+\frac{\gamma X}{\sigma}\geq u^{-\gamma}\right] \\ &=1-F_{\gamma, \sigma}\left[\frac{\sigma}{\gamma}\left( u^{-\gamma}-1 \right) \right]\\ &=u. \end{align*}
If $\gamma<0$, then \begin{align*} F_U(u) = \mathbb P(U\leq u) = \mathbb P \left[\left( 1+\frac{\gamma X}{\sigma} \right)^{-\frac{1}{\gamma}}\leq u\right] &=\mathbb P \left[1+\frac{\gamma X}{\sigma}\leq u^{-\gamma}\right] \\ &=\mathbb P \left[\frac{\gamma X}{\sigma}\leq u^{-\gamma}-1\right] \\ &=\mathbb P \left[X\geq\frac\sigma\gamma\left(u^{-\gamma}-1\right)\right] \\ &=1-F_{\gamma, \sigma}\left[\frac{\sigma}{\gamma}\left( u^{-\gamma}-1 \right) \right]\\ &=u. \end{align*}
If $\gamma=0$, then $F_{\gamma, \sigma}(x)=1-{\mathrm e}^{-x/\sigma}$, and it is easy to check that $\exp\left(-\frac1\sigma X\right)$ is uniformly distributed.